A396254 - OEIS

%I #14 May 21 2026 07:53:50

%S 1,1,2,1,8,2,8,1,12,6,2,6,12,1,40,20,40,20,2,20,40,20,40,1,48,12,8,6,

%T 48,2,48,6,8,12,48,1,128,32,128,8,128,32,128,2,128,32,128,8,128,32,

%U 128,1,108,54,12,54,108,6,108,54,2,54,108,6,108,54,12,54,108,1,220,110,220,110,220,110,220,110,220,110,2,110,220,110,220,110,220,110,220,110,220

%N Irregular table read by rows: let p = prime(n) and chi be any Dirichlet character modulo p of order p-1. Then T(n,k), 0 <= k <= p-2, is the algebraic degree of tau(chi^k) over Q, where tau is the Gauss sum.

%C If chi has order d, then Q(tau(chi)) is the compositum of Q(zeta_d) and Q(zeta_p)^G, where G is the unique index-d subgroup of (Z/pZ)*. See the Math Stack Exchange link below.

%H Jianing Song, <a href="/A396254/b396254.txt">Table of n, a(n) for n = 1..10123</a> (Rows 1..68)

%H Math Stack Exchange, <a href="https://math.stackexchange.com/questions/5137599/let-chi-be-a-nontrivial-dirichlet-charcater-of-order-d-modulo-p-is-it-tr">Let chi be a nontrivial Dirichlet charcater of order d modulo p. Is it true that zeta_d is in Q(tau(chi))?</a>.

%F T(n,k) = d*EulerPhi(d), where d = (prime(n)-1) / gcd(k, prime(n)-1).

%e The first rows:

%e   1;

%e   1, 2;

%e   1, 8, 2, 8;

%e   1, 12, 6, 2, 6, 12;

%e   1, 40, 20, 40, 20, 2, 20, 40, 20, 40;

%e   1, 48, 12, 8, 6, 48, 2, 48, 6, 8, 12, 48;

%e   1, 128, 32, 128, 8, 128, 32, 128, 2, 128, 32, 128, 8, 128, 32, 128;

%e   1, 108, 54, 12, 54, 108, 6, 108, 54, 2, 54, 108, 6, 108, 54, 12, 54, 108;

%e   1, 220, 110, 220, 110, 220, 110, 220, 110, 220, 110, 2, 110, 220, 110, 220, 110, 220, 110, 220, 110, 220;

%e   1, 336, 84, 336, 42, 336, 84, 8, 42, 336, 84, 336, 42, 336, 2, 336, 42, 336, 84, 336, 42, 8, 84, 336, 42, 336, 84, 336.

%e Let n = 3, prime(3) = 5. Let chi = [1->1, 2->i, 3->-i, 4->-1] be a Dirichlet character modulo 5 of order 4.

%e - tau(chi^0) = 0, which has degree 1, so T(3,0) = 1;

%e - tau(chi^1) = 2*sin(Pi/5) + 2*i*sin(2*Pi/5), which has degree 8 (in fact Q(2*sin(Pi/5) + 2*i*sin(2*Pi/5)) = Q(zeta_20)), so T(3,1) = 8;

%e - tau(chi^2) = sqrt(5), which has degree 8, so T(3,2) = 2;

%e - tau(chi^3) = -2*sin(Pi/5) + 2*i*sin(2*Pi/5), which has degree 8 (in fact Q(-2*sin(Pi/5) + 2*i*sin(2*Pi/5)) = Q(zeta_20)), so T(3,3) = 8.

%e Let n = 4, prime(4) = 7. Let chi = [1->1, 2->w^2, 3->w, 4->w^4, 5->w^5, 6->-1] be a Dirichlet character modulo 7 of order 6, where w = zeta_6.

%e - tau(chi^0) = 0, which has degree 1, so T(4,0) = 1;

%e - tau(chi^1) is a root to x^12 - 497*x^6 + 117649 = 0, which has degree 12 (in fact Q(tau(chi^1)) = Q(zeta_42)), so T(4,1) = 12;

%e - tau(chi^2) is a root to x^6 - 7*x^3 + 343 = 0, which has degree 6 (in fact Q(tau(chi^2)) = Q(zeta_42+zeta_42^{-1})), so T(4,2) = 6;

%e - tau(chi^3) = sqrt(-7), which has degree 2, so T(4,3) = 2;

%e - tau(chi^4) and tau(chi^2) have the same minimal polynomial, so T(4,4) = 6;

%e - tau(chi^5) and tau(chi^1) have the same minimal polynomial, so T(4,5) = 12.

%o (PARI) row(n) = my(p=prime(n), d); vector(p-1, i, d=(p-1)/gcd(i-1,p-1); d*eulerphi(d))

%Y Cf. A006093 (row lengths), A002618 ({d*EulerPhi(d)}), A104039 (column k=1).

%Y Cf. A396255 (another version).

%K nonn,tabf,easy

%O 1,3

%A _Jianing Song_, May 20 2026